Characterization of fields via equations 
A field $F$ is a commutative ring with identity and with at least two elements such that for all $a$ not equal to $0$ and $b$ in $F$, the equation $ax= b$ has a unique solution in $F$.

Here is what I did, but it's wrong.
I will denote the identity element of $G$ by $e$.
Fix any $a$ and $b$ in $G$. 
From the group axioms we know there is something called $a^{-1}$ in $G$ with the property that $a^{-1} * a = a * a^{-} = e$. Let $x = a^{-1} * b$;
this is an element of $G$ because of the group axioms (it comes from that $a^{-1}$ and $b$ are in $G$, and the fact that $*$ is a binary operation on $G$). than we get
$a * x = a * (a^{-1} * b)$ [from the definition of $x$]
$= (a * a^{-1}) * b$ [by the associative law]
$= e * b$ [by one of the properties of $a^{-1}$]
$= b$ [by the defining property of the identity element $e$].
This shows that there is indeed an element $x$ of $G$, namely $x = a^{-1} * b$, with the property that $a * x = b$. Since $a$ and $b$ were arbitrary elements of $G$.
 A: You don't say exactly what $G$ is, but it seems like you're trying to construct an inverse in the additive group underlying your field - you don't know that $F^\times$ forms a group yet, since this is what you're trying to prove (where $F^\times$ is the set of nonzero field elements). Hence, it appears that you're assuming your conclusion.
Recall that a field is a commutative ring with $1$ (which we have) such that every nonzero element has a multiplicative inverse. So all we need to do is to select a nonzero field element and give an inverse. Select $a \neq 0 \in F$. Because $1 \neq 0$, our assumption shows that
$$ax = 1$$
has a unique solution $x_0$; that is, $x_0$ is a right inverse for $a$. But we're in a commutative ring, so
$$x_0a = ax_0 = 1$$
so $x_0$ is also a left inverse for $a$. It follows that $a$ has a multiplicative inverse, completing the proof.
A: I think you called $\,G\,$ what at first you called $\,R\,$....anyway: $\,G\,$ indeed is a group with respect to addition , so multiplicatively we have nothing but the bare fact that $\,G\,$ is a unitary semigroup...
What's left to be proved is that multiplicatively $\,G^*:=G\setminus\{0\}\,$ is a group, which ammounts to prove that every non-zero element has a multiplicative inverse, and that's exactly what the given data tells you: let $\,0\ne a\in G\;$ , then there exists a unique
$$x\in G\;\;s.t.\;\;ax=1\implies x\;\;\text{is the wanted multiplicative inverse of}\;\;a$$
